Question

If in a $∆\mathrm{ABC}$, $A={\tan }^{-1}2$ and $B={\tan }^{-1}3$, then angle $C$ is

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. None of these.

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option.

Find the angle $C$:

Given,

$A={\tan }^{-1}2$ and $B={\tan }^{-1}3$.

In a triangle $\angle A+\angle B+\angle C=\mathrm{\pi }$.

$$\begin{gathered} \angle A+\angle B+\angle C=\mathrm{\pi } \\ \begin{array}{rcl}\Rightarrow {\tan }^{-1}2+{\tan }^{-1}3+\angle \mathrm{C}& =& \mathrm{\pi }\\ \Rightarrow {\tan }^{-1}\left(\frac{2+3}{1-2\times 3}\right)+\angle \mathrm{C}& =& \mathrm{\pi }\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{=}\mathbf{\pi }\mathbf{-}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{x}\mathbf{+}\mathbf{y}}{\mathbf{1}\mathbf{-}\mathbf{xy}}\mathbf{,}\mathbf{xy}\mathbf{>}\mathbf{1}\mathbf{]}\\ \Rightarrow \mathrm{\pi }-{\tan }^{-1}\left(\frac{5}{-5}\right)+\angle \mathrm{C}& =& \mathrm{\pi }\\ \Rightarrow \mathrm{\pi }-{\tan }^{-1}\left(-1\right)+\angle \mathrm{C}& =& \mathrm{\pi }\\ \Rightarrow \mathrm{\pi }-\frac{\mathrm{\pi }}{4}+\angle \mathrm{C}& =& \mathrm{\pi }\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(-1\right)\mathbf{=}\frac{\mathbf{-}\mathbf{\pi }}{\mathbf{4}}\mathbf{]}\\ \Rightarrow \angle \mathrm{C}& =& \frac{\mathrm{\pi }}{4}\end{array} \end{gathered}$$

Hence, the correct option is C.